How to find the value of an unknown exponent?

Solution 1:

... and without logarithms or knowing any powers of $2$ other than the most trivial one ... \begin{align*} 2^{4x+1} &= 128 \\ 2^{4x+1-1} = 2^{4x} &= 128/2 = 64 \\ 2^{4x-1} &= 32 \\ 2^{4x-2} &= 16 \\ 2^{4x-3} &= 8 \\ 2^{4x-4} &= 4 \\ 2^{4x-5} &= 2 \\ 2^{4x-6} &= 1 = 2^0 \text{,} \\ \end{align*} so $4x-6 = 0$ and $x = 6/4 = 3/2$.

Solution 2:

$2^{4x+1}=128\iff$

$\log_22^{4x+1}=\log_2128\iff$

${4x+1}=7\iff$

${4x}=6\iff$

${x}=6/4$

Solution 3:

$2^{4x+1}=128$

$\log 2^{4x+1}=\log 128$

$(4x + 1) \times \log 2 = \log 128$ - from properties of logs

$x = \frac{1}{4}(\frac {\log 128}{log (2)} - 1) = 3/2$

note that you can use any logarithm, log base 10 or 'ln' - or any other 'base' of logarithms you might have (with log10 and loge being the commonly found ones on calculators, spreadsheets etc ) you have to use your chosen type of log consistently of course

Solution 4:

To solve your equation, take the base-2 logarithm of both sides:

$$ \log_2 2^{4x+1} = \log_2 128 $$ $$ 4x+1 = 7 $$ $$ x = 1.5$$

However, is there a way of solving this without knowing that $128=2^7$?

Well, this piece of information is equivalent to “knowing that $\log_2 128 = 7$”, so no.

Solution 5:

It's very easy. Let's assume we have

$$k_1 ^ {k_2 * x + k_3} = k_4 $$

Where all the $k$'s are knowns, and the x is unknown.

We begin by removing the $k_3$ from the result ($k_4$) by dividing it by $k_1 ^ {k_3}$. The result thus obtained should be equal to:

$$k_1 ^ {k_2 * x}$$

As you can see we no longer have to deal with the $k_3$. Next we raise the result above to the power of $\frac{1}{k_2}$. That is, we find out what number we have to raise to the power of $k_2$ to reach at the result $k_1 ^ {k_2 * x}$. The answer should be equal to:

$$k_1 ^ x$$

Now finally from this result we can find out the value of $x$ by using a mathematical operation that lets us know, what number you have to raise a known other number to, to reach at a known result.

This operation is the logaritm. It's method of use is as follows: Let's assume $a ^ b = c$. The logarithm can be used to find out $b$, and it is then written as $\log_a c$.

So, to find out x in the result above, we use $\log_{k_1} result$ where $result$ is the value of the result obtained from previous operations.